Integrand size = 28, antiderivative size = 119 \[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\frac {2 a^2 (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}} \]
2*a^2*(c-c*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(a+a*sec(f*x+e))^(1/2)+2*a^2 *hypergeom([1, 1/2+n],[3/2+n],1-sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/ f/(1+2*n)/(a+a*sec(f*x+e))^(1/2)
Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=-\frac {2 a^2 \left (-6-4 n+(1+2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+n,\frac {5}{2}+n,1-\sec (e+f x)\right ) (-1+\sec (e+f x))\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt {a (1+\sec (e+f x))}} \]
(-2*a^2*(-6 - 4*n + (1 + 2*n)*Hypergeometric2F1[1, 3/2 + n, 5/2 + n, 1 - S ec[e + f*x]]*(-1 + Sec[e + f*x]))*(c - c*Sec[e + f*x])^n*Tan[e + f*x])/(f* (1 + 2*n)*(3 + 2*n)*Sqrt[a*(1 + Sec[e + f*x])])
Time = 0.45 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 4397, 3042, 4400, 75}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a \sec (e+f x)+a)^{3/2} (c-c \sec (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 4397 |
\(\displaystyle a \int \sqrt {\sec (e+f x) a+a} (c-c \sec (e+f x))^ndx+\frac {2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx+\frac {2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 4400 |
\(\displaystyle \frac {2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}-\frac {a^2 c \tan (e+f x) \int \cos (e+f x) (c-c \sec (e+f x))^{n-\frac {1}{2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (1,n+\frac {1}{2},n+\frac {3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}\) |
(2*a^2*(c - c*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]] *(c - c*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x] ])
3.2.39.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[-2*a^2*Cot[e + f*x]*((c + d*Csc[e + f *x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a Int[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LeQ[n, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
\[\int \left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}} \left (c -c \sec \left (f x +e \right )\right )^{n}d x\]
\[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\text {Timed out} \]
\[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]